Sturm–Liouville Eigenvalue Problem: 04-BS-5, December 2019, Q1

Question 1: Sturm–Liouville Eigenvalue Problem (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The self-adjoint (Sturm–Liouville) equation $(x^{-3}y')'+(\lambda+4)x^{-5}y=0$ on $1\le x\le e^2$, with $p(x)=x^{-3}$, weight $w(x)=x^{-5}$, and homogeneous boundary conditions $y(1)=0,\ y(e^2)=0$.

Find. All eigenvalues $\lambda_n$ and the corresponding eigenfunctions $y_n(x)$.

Approach. Multiply through by $x^5$ to collapse the SL form into a Cauchy–Euler equation, solve its characteristic equation, discard the cases that cannot satisfy two homogeneous end conditions, then apply the boundary conditions to the surviving oscillatory family.

Reduce to Cauchy–Euler form. Expanding the derivative, $(x^{-3}y')'=x^{-3}y''-3x^{-4}y'$, so the equation reads $$x^{-3}y''-3x^{-4}y'+(\lambda+4)x^{-5}y=0$$ Multiplying every term by $x^{5}$ clears the negative powers: $$\boxed{x^{2}y''-3xy'+(\lambda+4)y=0}$$ a linear Cauchy–Euler (equidimensional) equation.
Solve the characteristic equation. Trying $y=x^{m}$ gives $m(m-1)x^m-3mx^m+(\lambda+4)x^m=0$, so $$m^{2}-4m+(\lambda+4)=0\ \Rightarrow\ m=2\pm\sqrt{4-(\lambda+4)}=2\pm\sqrt{-\lambda}$$ For $\lambda\le0$, $m$ is real and $y=C_1x^{m_1}+C_2x^{m_2}$; enforcing $y(1)=0$ and $y(e^2)=0$ on such a solution forces $C_1=C_2=0$ (only the trivial solution), so no non-positive $\lambda$ is an eigenvalue — the eigenvalues must be positive.
Positive $\lambda$: oscillatory solution. Write $\lambda=\mu^2$ ($\mu\gt0$), so $m=2\pm i\mu$ and $$y(x)=x^{2}\big[C_1\cos(\mu\ln x)+C_2\sin(\mu\ln x)\big]$$
Apply $y(1)=0$. Since $\ln 1=0$, $y(1)=1^2[C_1\cos0+C_2\sin0]=C_1=0$.
Apply $y(e^2)=0$. With $C_1=0$ and $\ln(e^2)=2$, $$y(e^2)=e^{4}\,C_2\sin(2\mu)=0$$ For a non-trivial eigenfunction $C_2\ne0$, so $\sin(2\mu)=0\Rightarrow 2\mu=n\pi,\ n=1,2,3,\dots$
Assemble the eigenpairs. $\mu_n=n\pi/2$, hence $$\boxed{\lambda_n=\dfrac{n^2\pi^2}{4},\qquad y_n(x)=x^{2}\sin\!\Big(\dfrac{n\pi}{2}\ln x\Big),\qquad n=1,2,3,\dots}$$ The first three numerical eigenvalues are $\lambda_1=\pi^2/4\approx2.467401$, $\lambda_2=\pi^2\approx9.869604$, $\lambda_3=9\pi^2/4\approx22.206610$.

Quantity	Result
Reduced ODE	$x^2y''-3xy'+(\lambda+4)y=0$ (Cauchy–Euler)
Eigenvalues	$\lambda_n=n^2\pi^2/4,\ n=1,2,3,\dots$
Eigenfunctions	$y_n(x)=x^2\sin\!\left(\dfrac{n\pi}{2}\ln x\right)$
$\lambda_1,\lambda_2,\lambda_3$	$2.467401,\ 9.869604,\ 22.206610$

Topic: Regular Sturm–Liouville eigenproblem on a Cauchy–Euler (equidimensional) operator.

Key relations: SL self-adjoint form $(py')'+(q+\lambda w)y=0$; Cauchy–Euler trial solution $y=x^m$; characteristic equation $m^2+(p_0-1)m+q_0=0$.

Why this works: multiplying by $x^5$ converts the singular-looking $x^{-3},x^{-5}$ coefficients into the classical equidimensional form, whose $y=x^m$ ansatz always reduces the ODE to an algebraic equation in $m$. Testing $\lambda\le0$ first (and showing it forces the trivial solution) proves the eigenvalues are strictly positive before spending effort on the oscillatory case; this mirrors the general SL theorem that eigenvalues of a problem with this sign structure are bounded below.

Common pitfall: forgetting to rule out $\lambda\le0$ and instead assuming the oscillatory root form applies for all $\lambda$; also mis-evaluating $\ln(e^2)=2$ as $e^2$ when applying the second boundary condition.

Source: Kreyszig, Advanced Engineering Mathematics, 10th ed., Ch. 11.5 (Sturm–Liouville Problems); Ch. 2.5 (Euler–Cauchy equations).

Question 1 of 7: Sturm–Liouville Eigenvalue Problem

Question 1: Sturm–Liouville Eigenvalue Problem (20 marks)